Chem 1412 Spring 2016 – Daily Quiz #25
Name
Potentially Useful Information
x=
−b± b 2−4ac
2a
Questions
1. Carbonic acid (H2CO3) is a diprotic acid. The Kas are as follows: Ka1 = 4.3 x 10­7
Ka2 = 4.7 x 10­11
What is the pH of a 0.222 M solution of phosphoric acid?
A. 6.37
B. 7.92
C. 3.57
D. 2.46
E. 6.53
RICE table
Reaction
Initial
Change
Equilibirum
A. Piperidine
B. Piperazine
C. They are the
same. piperidine piperazine
H2CO3 ⇌
2.22
-x
2.22 -x
H
0
+x
+x
+
HCO
0
+x
+x
–
3
You will ignore Ka2 because it is much smaller than Ka1.
4.3×10−7 =
2. Which of these compounds is the stronger base (the one more likely to become protonated?) x2
0.222−x
assume x is small
x = √ (4.3×10−7 )(0.222) = 3.0 9×10−4
pH = -log(3.09 x 10-4) = 3.57
The electronegativity of the extra nitrogen in piperazine pulls electrons away from the
hydrogens. This makes the nitrogens less likely to accept a proton. piperidine pKb = 2.79
piperazine pKb = 4.19
Worksheet 25
3. Which of the following will produce an acidic solution in water? X. NH4NO3
Y. NaCl
Z. NaF
5. Which of the following will produce an neutral solution in water? X. NH4NO3
Y. NaCl
Z. NaF
A. X and Y
B. X
C. Y
D. Z
E. Y and Z
A. X and Y
B. X
C. Y
D. Z
E. Y and Z
Na+ and Cl– are derived from a strong base (NaOH) and a strong acid (HCl) and so are neutral in solution. See the answer to question 3.
NH4NO3 produces NO3– which is derived from a strong acid and therefore neutral and NH4+, which makes an acidic solution: NH4+ + H2O ⇌ H3O+ + NH3
6. If you add 0.1 M NaNO3 to 0.1 M HNO3 (see the reaction below) will the pH change? HNO3(aq) + H2O(ℓ) ⇌ H3O+(aq) + NO3–(aq)
A. No, because neither Na+ or NO3– will change the pH. B. No, because of Le Châtliers Principle.
C. Yes, because Na+ will change the pH.
D. Yes, because of Le Ch â
tliers Principle.
–
E. Yes, because NO3 will change the pH. 4. Which of the following will produce an basic solution in water? X. NH4NO3
Y. NaCl
Z. NaF
When you add NO3–, you are adding a product, which, by Le Châtliers Principle, will cause the reaction to shift toward reactants. Here, this causes a reduction in the [H3O+], which raises the pH. This is called the common ion effect (NO3– is the common ion.)
A. X and Y
B. X
C. Y
D. Z
E. Y and Z
F– + H2O ⇌ HF + OH–
Page 2
Worksheet 25
7. If you add 0.3 M NaHSO3 to a solution of 0.1 M H2SO3 (see reaction), how will the pH change? H2SO3(aq) + H2O(aq) ⇌
H3O+(aq) + HSO3–(aq)
A. It will stay the same.
B. It will go down.
C. It will go up.
By adding HSO3–, you are adding a product, which, Le Châtliers Principle, shifts the reaction toward the reactants. This decreases the concentration of H3O+ so the pH will go up. This is also the common ion effect. 8. Consider the acid base reaction below and
choose the base­conjugate acid pair from the list. CH3NH2(aq) + H2O(ℓ) ⇌
CH3NH3+(aq) + OH–(aq)
base conjugate acid A. CH3NH2 CH3NH3+
B. CH3NH2
OH– C. H2O
OH– D. H2O
CH3NH3+ E. None of the above. CH3NH2 accepts a proton so it is the base (Brønsted­Lowry). CH3NH3+ is derived from
the base, so it is the conjugate acid. Page 3
CHEM 1412 MWF Spring 2016
Worksheet #25 – pH of Salts and the Common Ion Effect – Key
Name Team pH of Salts and the Common Ion Effect
Why?
The addition of some salts to water causes a change in pH. Other salts don't cause a change
in pH. Which salts cause changes to the pH and does the pH go up or down? How can you
calculate the pH? What happens to the pH when you add a neutral salt, such as NaNO 2 to a
dilute solution of HNO2? Stay tuned. Learning Objectives
Students should be able to: ● Predict whether a salt solution will be acidic or basic. ● Calculate the pH of salt solutions.
● Predict whether the addition of a salt to an acid­base equilibrium will cause the pH of a solution to increase or decrease. ● Calculate the pH of solutions in the presence of added salt. Resources
Gilbert, 16.6 and 16.7
ChemTours
No ChemTours today.
Videos
pH of Salts
https://vimeo.com/20873043
This video summarizes the basic pH effects of salts. 10:12 minutes. Calculation of the pH of 0.1 M Sulfuric Acid
https://www.youtube.com/watch?v=6exH6k9k60o
This video works out the problem presented in the book on p 797. 12:11 minutes
Prerequisites
Equilibrium, algebra, Ka, Kb, Kw, conjugate acid, and conjugate base. Worksheet 25
Vocabulary
Common ion effect. Focus Information
Common Ion Effect
Good news! If you have been paying attention, you already know this! The common ion effect is just Le Chatlier's principle! In the common ion effect, the addition oft salts that have a common ion with a weak acid can change the pH. For instance, consider what happens when
you add NaF to the reaction below. Good news! If you have been paying attention, you already know this! The common ion effect is
just Le Chatlier's principle! In the common ion effect, he addition oft salts that have a common
ion with a weak acid can change the pH. For instance, consider what happens when you add
NaF to the reaction below.
HF(aq) ⇌ H+(aq) + F­(aq)
That was easy! By the way, when the [H+] goes down, the pH goes up. Key Questions
1. What is the pH of a 0.5 M solution of NaCl? Answer pH 7.0
Salts that consist of the cations of strong bases and the anions of strong acids have no effect on pH when dissolved in water. Page 2
Worksheet 25
2. Are solutions of the following salts acidic, basic or neutral? For acidic or basic solutions write the appropriate chemical equation. a. NaNO2
Basic
b. NaNO3
Neutral
c. C5H5NHClO4 Acidic NO2­ + H2O ⇌ HNO2 + OH­
C5H5NH+ ⇌ C5H5N­ + H+
d. NH4NO2
Acidic
NH4+ ⇌ NH3 + H+
Note NH4+ pKa = 9.25, NO2– pKb = 10.85
NH4+ is a stronger acid than NO2­ is a base. pKa < pKb
e. KOCl
Basic
OCl­ + H2O ⇌ HOCl + OH­
f. CO2 Acidic
CO2 + H2O ⇌ H2CO3
g. SO3 Acidic
SO3 + H2O ? H2SO4
3. Arrange the following 0.10 M solutions in order of most acidic to most basic. KOH, KNO3, KCN, NH4Cl, HCl
Answer HCl > NH4Cl > KNO3 > KCN > KOH HCl = strong acid
NH4Cl – NH4+ is the conjugate acid of a weak base – produces an acid solution
KNO3 – NO3– is the conjugate base of a strong acid – this is neutral KCN – CN– is the conjugate base of a weak acid – produces a basic solution
KOH – strong base Page 3
Worksheet 25
4. a. What are Kb and pKb for H2BO3–?
H3BO3(aq) ⇌ H2BO3–(aq) + H+(aq)
­5
Kb = 1.9 x 10
Ka = 5.4 x 10­10
pKb = 4.72
Kw
1×10−14
Kb =
=
= 1.9×10−5
−10
Ka
5.4×10
pKb = ­log(1.9 x 10­5) = 4.72 b. What are Ka and pKa for (CH3)2NH2+?
(CH3)2NH(aq) + H2O(ℓ) ⇌ (CH3)2NH2+(aq) + OH–(aq) Kb = 1.7 x 10
­11 Kb =
Kw
=
Ka
1×10−14
= 1.7×10−11
−4
5.9×10
Kb = 5.9 x 10­4
pKb = 10.77
pKb = ­log(1.7 x 10­11) = 10.77
5. Calculate the pH of a 0.10 M solution of NaOCl. The pK a of HOCl is 3.5 x 10­8.
Answer 10.23
The reaction is: OCl­ + H2 O ? HOCl + OH­ You need to calculate the Kb from the Ka:
­
Kb for OCl = Kw
Ka
=
1.0 x10−14
= 2. 86×10−7
−8
3.5 x10
RICE table
Reaction
Initial
Change
Equilibirum
OCl–(ℓ) + H2O(ℓ) ⇌
0.10
-x
0.10-x
Page 4
HOCl OH 0
0
+x
+x
x
x
Worksheet 25
Kb =
 xx
assume x is small
0.1−x
2. 86x10 −7 =
x2
0.1
check x is small x = √ 2.86x10−8 = 1. 69x10−4 1.6 9 x 10−4
× 100 = 0.17 % so x is small!
0.1
[OH­] = 1.69 x 10­4 pOH = 3.77
pH = 14.00 – 3.77 = 10.23
6. Calculate the pH of a 0.25 M solution of methylamine chloride (CH 3NH3Cl) in water. The Kb for methylamine (CH3NH2) is 4.38 x 10­4.
Answer 5.622
methylamine chloride disassociates in water to CH 3NH3+ and Cl–. The chloride ion has no affinity for protons so the equilibrium we are interested in is: CH3NH3+ ⇌ CH3NH2 + H+ so the equilibrium we want is: Ka =
[CH3 NH 2 ][H+ ]
[CH3 NH+3 ]
so we need to calculate Ka from the Kb that we have:
K w = Ka ×Kb and Ka =
Kw
kb
=
1.00 x 10−14
−4
4.38 x 10
−11
= 2.2 8 3×10
RICE table
Reaction
Initial
Change
Equilibirum
CH3NH3+ ⇌
0.25
-x
0.25-x
Page 5
CH 3NH2 H+
0
0
+x
+x
x
x
Worksheet 25
−11
Ka = 2.2 8 3×10
x2
=
assuming x is small,
0.25−x
x = √ 0.25×2.2 8 3×10−11 = 2.3 8 9 x 10−6
2.389 x 10−6
×100 = 0.0010 Yes!
check x is small: 0.25
check x is correct: Ka = 2.2 8 3×10−11 =
(2.389×10−6 )2
−11
= 2.2829 x10
0.25
Yes!
calculate the pH: = ­log [H+] `=` ­log (2.389 x 10­6) = 5.6218 = 5.622
7. For each case below indicate whether the equilibrium shifts to the right or to the left with the addition of the compound indicated and say whether the pH increases or decreases after you add the compound. a. HF(aq) ⇌ H+(aq) + F–(aq) Reaction shifts left add BaF2(aq)
pH goes up
b. NH3(aq) + H2O(l) ⇌ NH4+ (aq) + OH–(aq)
Reaction shifts left pH goes down
c. CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH–(aq)
methylamine
Reaction shifts left
add CH3NH3Cl(aq)
pH goes down
d. CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO–(aq) acetic acid
acetate
Reaction shifts left add NH4NO3
pH goes up
Page 6
add CH3COONa(aq)
(soduim acetate)
Worksheet 25
8. You make two solutions Solution A = 0.1 M NH4Cl
Solution B = 0.1 M NH4Cl(aq) + 0.1 M NH3(aq)
a. Will solution A be acidic or basic? NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)
Answer acidic
b. Which solution will have the lower pH?
Answer solution A
Solution A: NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
In solution B, the addition of extra NH3 drives the equilibrium to the left, thus lowering the [H+] and raising the pH. 9. Calculate the pH of a solution containing 0.20 M HC 2H3O2 (acetic acid) Ka 1.8 x 10­5 and 0.50 M NaC2H3O2 (sodium acetate). Answer pH = 5.14
HC2H3O2 ⇌ C2H3O2– + H+
RICE table
Reaction
Initial
Change
Equilibirum
HC2H 3O2 ⇌
0.2
-x
0.2-x
−5
Ka = 1.8×10
=
[C2H3O2-]
0.5
+x
0.5-x
[H+ ][C2 H3 O­2 ]
[HC2 H 3 O 2 ]
=
[H +]
0
+x
+x
Note: [C2H3O2–]0 = 0.5 M because the NaC2H3O2 dissociates in solution. (x)(0.50+ x) (x)(0.50)
=
0.20−x
(0.20)
(assuming x is small in both cases) Solving for x,
1.8×10−5 0.2 
7.2×10−6
x=
= 7.2 x 10−6 check x is small: ×100 % = 3.6×10−3 %
0.5
0.20
pH = ­log(7.2 x 10­6) = 5.14
Page 7